The question is using method of characteristics to find the integral surface of $pq=xy$ with initial data curve $x=1$, $y=z^2$.
I have tried every possible way but was far away from the result. The answer i got is $$x=\cosh t + \frac1s\sinh t ,$$ $$y=2s^3\cosh t +\sinh t ,$$ $$z=\frac{2s^3+s}{2}\cosh 2t+\frac{\sinh 2t+2t}{2} 3s^3-2s^2 t+\frac{s-2s^3}{2} .$$ The problem remains is to eliminate $s$ and $t$ from above equation so we get integral surface and i am helpless